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The small-argument Taylor series of the single-site response isLet $K=4\beta J$. For a small nonzero order parameter at zero field, the self-consistency equation becomesSince $K_c=3/2$ and $K=K_c/(1+t)$, where $t=T/T_c-1$, this givesConsequentlyThe positive branch is the expression printed in the question, while its negative partner is required by spin inversion symmetry. The square-root behavior gives the mean-field critical exponent $1/2$ for the spontaneous magnetization.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]

Take $a=a_0(T-T_c)$ and hold $b,c$ fixed. The heat capacity comes from the minimized free energy, $C=-T{d}^2{F}_{\min }/d{T}^2$, with any smooth background added separately. For $b>0$ and $T<T_c$,Above $T_c$, $m=0$ and the polynomial contributes zero. There is a finite jump, rather than a divergence: the conventional ordinary mean-field critical exponents areThe upper branch has zero order-parameter contribution; a finite regular heat capacity has exponent zero.

Along the tuned tricritical path $b=0$, the ordered minimum givesUpper-side qualification: the vanishing disordered-side heat-capacity amplitude in Landau theory matters here: for $b=0$ and $a>0$, the minimizing solution is still exactly $m=0$, so this saddle-point polynomial gives $F_{\min }-F_{\mathrm{reg}}=0$ and $C_{\mathrm{sing},+}=0$. A zero function cannot determine a nonzero-amplitude power-law exponent. For example, adding the allowed smooth background $F_{\mathrm{reg}}=-C_{0}T\log (T/T_{\ast })$, with $C_0>0$, leaves the entire phase diagram unchanged but makes the upper-side heat capacity exactly $C_0$, not a $(T-T_c{)}^{-1/2}$ divergence.

One convention assigns the tricritical scaling exponent $\alpha =1/2$ to both sides, writing $C_{\mathrm{sing},\pm }=A_{\pm }|T-T_c{|}^{-1/2}$ with $A_{+}=0$ at this mean-field approximation saddle. If the source's literal $C\sim |T-T_c{|}^{-{\alpha }_{+}}$ is meant to have a nonzero leading coefficient, the given polynomial establishes no upper-side tricritical singular exponent; with a nonzero smooth background its leading upper-side exponent is zero. A nonzero upper-side singularity requires fluctuation information beyond this minimization. This distinction prevents claiming an upper-side divergence that the displayed Landau free energy does not produce.

For the stable sextic Landau free energy $f=r{m}^2/2+q{m}^4/4+s{m}^6/6-hm$, $s>0$, rescaling $m=(|r|/s{)}^{1/4}\psi$ gives the minimized order-parameter contribution $f_{\mathrm{eq}}=|r{|}^{3/2}{s}^{-1/2}{\Phi }_{\sigma }(X,Y)$, where $\sigma =\mathrm{sgn}r$, $X=q/(|r|s{)}^{1/2}$ and $Y=h{s}^{1/4}/|r{|}^{5/4}$. The two functions are ${\Phi }_{\sigma }={\min }_{\psi }(\sigma {\psi }^2/2+X{\psi }^4/4+{\psi }^6/6-Y\psi )$. In particular ${\Phi }_{-}(0,0)=-1/3$, but ${\Phi }_{+}(0,0)=0$. These separate branches and removal of smooth background terms are necessary when using the homogeneous form to infer mean-field critical exponents.